Why is $2\pi i \neq 0?$

Question

We know that $e^{\pi i} = -1$ because of de Moivre's formula. ($e^{\pi i} = \cos \pi + i\sin \pi = -1).$

Suppose we square both sides and get $e^{2\pi i} = 1$(which you also get from de Moivre's formula), then shouldn't $2\pi i=0$? What am I missing here?

Answer 1

You have shown that $e^{2\pi i} = e^0$. This does not imply $2\pi i = 0$, because $e^z$ is not injective. You have to give up your intuition about real functions when you move them to the complex plane, because they change a lot. $e^z$ is actually periodic for complex $z$.

Answer 2

The $\log$ function is multi-valued on $\mathbb{C}^*$ (you can however choose a "branch" of it; see Wikipedia). At any rate, just because $$e^{2\pi i}=e^0$$ does not imply that $2\pi i=0$.

Answer 3

It's like saying that, because $\sin{\pi} = \sin{0}$, that $\pi = 0$. Not all functions have perfect inverses; sin being one of them. In the complex numbers, $e^z$ doesn't either.

You're implicitly going: $e^{2\pi i} = e^0 \implies \ln{e^{2\pi i}} = \ln{e^0} \implies 2 \pi i = 0$. The error here is that $\forall x \in \mathbb{C} \ \ln{e^x} = x$ is not true! $\ln$ isn't even a function, just like the naive version of $\arcsin(x)$. You have to make a choice of range, which is usually $\Im(x) \in (-\pi, \pi]$.

Answer 4

Actually, the following is true:

$ e^{2n\pi i} = 1  \forall  n \in \mathbb{Z} $

As a special case of $n = 0$ this gives what you know: $e^0 = 1$, but as mentioned, this is only a special case of the general formula given above.

You could as well assume that $2\pi = 0$ because of

$\sin 2\pi = \sin 0 \land \cos 2\pi = \cos 0$.

As others already mentioned, this stuff just isn't injective, so conclusions relying on the injectivity can be wrong.